The terminals of a battery of emf 24.0 V and internal resistance of 1.0 Ω is connected to an external resistor 5.0 Ω. Find the terminal p.d.
A.
18.0V
B.
12.0V
C.
16.0V
D.
20.0V
Correct answer is D
ε = 24.0 V, r=1Ω, R=5.0Ω
terminal p·d = ?
ε = IR+Ir = I(R+r)
⇒ IR+Ir = I(R+r)
⇒ I = \(\frac{24}{5+1}\)
⇒ I =\(\frac{24}{6}\)
⇒ I = 4 A
terminal p·d = 4 × 5
terminal p·d = 20.0 V
